Document 1807175

Chapter 6,
continued
4
1
3. } 5 }
16
a13
a
36. Let } 5 r.
b
a
c
e
Because r 5 }b 5 }d 5 }f , you know the following:
a
c
e
2. r 5 }d
1. r 5 }b
a 5 br
3. r 5 }f
c 5 dr
e 5 fr
br 1 dr 1 fr
r(b 1 d 1 f )
a
a1c1e
So, } 5 }
5}
5 r 5 }b .
b1d1f
(b 1 d 1 f )
b1d1f
37. Water content in 5 kg of fresh apricots: 5(0.86) 5 4.3 kg
Let x be the number of kilograms of water removed from
the fruit.
Water removed:
Water content after dehydration
75% 5 }}}
Weight after dehydration
3
4
4.3 2 x
52x
}5}
16 5 4(a 1 3)
48 5 4(d 2 6)
16 5 4a 1 12
48 5 4d 2 24
15a
18 5 d
5
x
9
9
5. If } 5 }, then } 5 }.
2
2
5
x
Property of Proportions (Property 3)
y
15
x
x
6. If } 5 }, then } 5 }.
21
21
15
y
Property of Proportions (Property 3)
y
y 1 12
x18
x
7. If } 5 }, then } 5 }.
12
12
8
8
Property of Proportions (Property 4)
x1y
x
32
37
8. If } 5 }, then } 5 }.
y
y
5
5
Property of Proportions (Property 4)
}
3(5 2 x) 5 4(4.3 2 x)
9. The midpoint of AD is B and AD 5 10,
15 2 3x 5 17.2 2 4x
so AB 5 BD 5 5.
x 5 2.2
}
AC 5 ÏAB + AD
So, about 2.2 kilograms of water are removed from the
fruit. The weight of the dehydrated apricots is about
5 2 2.2 5 2.8 kg.
Mixed Review for TAKS
}
5 Ï 5 + 10
}
5 Ï 50
}
5 5Ï 2
}
So, AC is 5Ï2 ø 7.1 units.
38. A;
Lesson 6.3
f (x) 5 25(x 1 3)
Investigating Geometry Activity 6.3 (p. 371)
y 5 25(x 1 3)
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
4
6
4. } 5 }
8
d26
y 5 25x 2 15
The y-intercept is 215.
39. F;
Total
Number of
Number
of student 1 non-student 5 number
of tickets
tickets
tickets
Total
Cost of
Number
Cost of
Number
dollar
1
nonof
non1 student + of student 1
5
+
amount
student
student
ticket
tickets
of tickets
ticket
tickets
The system that represents the situation is
s 1 n 5 535
Photo 1
Photo 2
Measurement
Photo 1
Photo 2
}
AB
35 mm
25 mm
1.4
AC
14 mm
10 mm
1.4
DE
28 mm
20 mm
1.4
mŽ 1
608
608
1
mŽ 2
288
288
1
1. When a figure is reduced, the ratios of each pair of
corresponding lengths are equal.
2. When a figure is reduced, corresponding angles are
congruent.
4s 1 5.5n 5 2455.
3. The corresponding angles are congruent, so the measure
Quiz 6.1– 6.2 (p. 370)
5
10
1. } 5 }
2
y
of the corresponding angle in Photo 1 is 358.
2.
x
6
9
3
}5}
20 5 5y
3x 5 54
45y
x 5 18
Length in Photo 1
1.4
4. }} 5 }
1
Length in Photo 2
x cm
1 cm
1.4
1
}5}
x 5 1.4
The measure of the corresponding segment in Photo 2 is
about 1.4 centimeters.
Geometry
Worked-Out Solution Key
161
Chapter 6,
continued
Length in Photo 1
1.4
5. }} 5 }
1
Length in Photo 2
5 cm
6.3 Exercises (pp. 376–379)
Skill Practice
1.4
1
}
x cm 5 }
1. Two polygons are similar if corresponding angles
5 5 1.4x
are congruent and corresponding side lengths are
proportional.
3.571 ø x
The measure of the corresponding segment in Photo 2 is
about 3.6 centimeters.
6.3 Guided Practice (pp. 372–375)
1. Ž J > Ž P, Ž K > Ž Q, Ž L > Ž R;
KL
QR
LJ
RP
TQ
5
6
8
1 QR
1 TS
1
2. } 5 } 5 }, } 5 } 5 }, } 5 } 5 }
10
2 AB
12
2 DC
16
2
DA
1
The scale factor of QRST to ABCD is }2 .
4. Ž D > Ž P, Ž E > Ž Q, Ž F > Ž R, Ž G > Ž S;
x
4
}5}
DE
PQ
4. Because ABCDE is similar to FGHJK, the scale factor is
15
3
5 }2 .
the ratio of the lengths, }
10
15
10
18
x
5. } 5 }
HJ
WX
JK
XY
KL
YZ
LH
ZW
}5}5}5}
BC
AB
6. D; n ABC , nDEF, so } 5 }.
DE
EF
The correct answer is D.
}5}
7. All angles are right angles, so corresponding angles are
congruent.
180 5 15x
12 5 x
6. Perimeter of FGHJK: 15 1 9 1 12 1 15 1 18 5 69
units
Use Theorem 6.1 to find the perimeter x of ABCDE.
69
3
}5}
x
2
}5}5}
RS
WX
64
32
2
1
}5}5}
TU
YZ
64
32
2
1
}5}5}
}5}5}
ST
XY
48
24
2
1
UR
ZW
48
24
2
1
The ratios are equal, so the corresponding side lengths
are proportional. So, RSTU , WXYZ. The scale factor
2
of RSTU to WXYZ is }1.
138 5 3x
8. You can see that Ž C > Ž T, Ž D > Ž U, Ž E > Ž V.
46 5 x
So, corresponding angles are congruent.
The perimeter of ABCDE is 46.
7. Scale factor of nJKL to nEFG:
CD
TU
10
8
5
4
} 5 } 5 },
5
4
DE
UV
} 5 },
EC
VT
12
9.6
120
96
5
4
}5}5}5}
}5}5}
The ratios are equal, so the corresponding side lengths
are proportional. So, nCDE , nTUV. The scale factor
Because the ratio of the lengths of the medians in similar
triangles is equal to the scale factor, you can write the
following proportion.
of nCDE , nTUV is }4.
JL
EG
96
80
KM
FH
6
5
x
35
6
5
6
5
}5}
}5}
x 5 42
}
The length of the median KM is 42.
162
GD
SP
5. Ž H > Ž W, Ž J > Ž X, Ž K > Ž Y, Ž L > Ž Z;
85x
KF
EA
FG
RS
EF
QR
}5}5}5}
64 5 8x
FG
AB
If two polygons are similar, then corresponding angles
are congruent and corresponding side lengths are
proportional. Because two proportional side lengths
are not always congruent, two similar polygons are not
always similar.
BC
CA
AB
3. Ž A > Ž L, Ž B > Ž M, Ž C > Ž N; } 5 } 5 }
LM
MN
NL
DC
BC
3. } 5 }
TS
RS
16
8
the corresponding angles are congruent and the
corresponding side lengths are congruent. The ratio
of the side lengths of congruent sides is 1 : 1, so the
corresponding side lengths are proportional. So, two
congruent polygons must be similar.
Geometry
Worked-Out Solution Key
5
JK
20
5
9. } 5 } 5 }
EF
8
2
5
The scale factor of JKLM to EFGH is }2.
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
JK
PQ
}5}5}
2. Yes; no; If two polygons are congruent, then
Chapter 6,
continued
JK
KL
10. Find x: } 5 }
EF
FG
Find y: }
5}
HE
EF
20
8
}5}
x
11
}5}
8x 5 220
MJ
JK
30
y
20
8
240 5 20y
x 5 27.5
12 5 y
isosceles triangle has at least two congruent sides. So, the
ratios of corresponding side lengths of a scalene triangle
and an isosceles triangle can never all be equal. So, a
scalene triangle and an isosceles triangle are never similar.
18. x : 1; The definition states that the “ratio of a to b is a : b.
Find z: Ž J > Ž E
You can determine that the “ratio of b to a” is b : a.
So, switch the order of the given ratio.
65 5 z
19. The special segment shown in blue is the altitude.
11. Perimeter of EFGH:
EF 1 FG 1 GH 1 HE 5 8 1 11 1 3 1 12 5 34
Perimeter of JKLM: Use Theorem 6.1 to find the
perimeter x.
27
18
x
16
}5}
432 5 18x
24 5 x
5
2
x
34
17. Never; A scalene triangle has no congruent sides and an
}5}
20. The special segment shown in blue is the median.
2x 5 170
18
y
x 5 85
The periemter of EFGH is 34 and the perimeter of JKLM
is 85.
18y 2 18 5 16y
2y 5 18
y59
12. Let x be the small sign’s perimeter.
60 in.
x in.
5
3
6
8
21. } 5 }
8
x
}5}
180 5 5x
6
8
36 5 x
6y 5 80
2
2
1
2
13 }3 inches.
Perimeter of A: 10 1 12 1 6 5 28
6
3
22. Scale factor: } 5 }
8
4
2
Perimeter of B: }
5 }1
x
4.8
x
x 5 14
19.2 5 3x
6.4 5 x
14. Sometimes;
A
8
5
3
4
}5}
The perimeter of B is 14.
5
y 5 13 }3
The other two sides of nRST are 10 }3 inches and
Scale factor of A to B: }
5 }1
5
28
1
x 5 10 }3
13. The scale factor was used incorrectly.
10
8
B
4
4
C
6
12
3
nB , nC, but n A ï nB.
15. Always; The angles of all equilateral triangles are
congruent, so corresponding angles are always
congruent. Because the sides of an equilateral triangle
are congruent, the ratios of corresponding side lengths
of two equilateral triangles are always congruent. So,
two equilateral triangles are always similar.
16. Sometimes;
The length of the corresponding altitude in nRST is
6.4 inches.
4
19 }5
BC
19.8
198
11
23. } 5 } 5 } 5 } 5 }
EF
5
9
9
90
11
The scale factor of n ABC to nDEF is }
.
5
AB
11
24. Find DE: } 5 }
DE
5
22
x
11
5
}5}
110 5 11x
6
6 2
10
y
}5}
6x 5 64
The small sign’s perimeter is 36 inches.
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
16
y21
}5}
8
D
8
E
4
4
F
6
4
AC
11
y
11
Find AC: }
5}
DF
5
}5}
2
5
10 }5
5y 5 114.4
10 5 x
y 5 22.88
}
}
The length of DE is 10 and the length of AC is 22.88.
4 2
nD , nF, but n D ï nE.
Geometry
Worked-Out Solution Key
163
Chapter 6,
25.
x
8
continued
11
5
}5}
Problem Solving
length of court
78 ft
26
31. }} 5 } 5 }
3
length of table
9 ft
5x 5 88
x 5 17.6
The length of the altitude shown in n ABC is 17.6.
1
26. Area of n ABC: A 5 } bh
2
1
5 }2 (22.88)(17.6)
5 201.344
width of court
width of table
36
5
The ratios are not equal, so the corresponding side
lengths are not proportional, and the surfaces are
not similar.
width of computer screen
13.25 in.
1
32. }} 5 } 5 }
4
width of projected image
53 in.
height of computer screen
height of projected image
10.6 in.
42.4 in.
1
4
}} 5 } 5 }
1
Area of n DEF: A 5 }2 bh
5 }2 1 10}5 2(8)
The ratios are equal, so the corresponding side lengths
are proportional, and the surfaces are similar. The scale
5 41.6
factor of the computer screen to the projected image is }4.
2
1
1
The ratio of the area of n ABC to n DEF is
201.344
41.6
33. a.
} 5 4.84, which is the square of the scale
15
36 ft
5 ft
}} 5 } 5 }
BC
4
CD
6
7
10
}5}
2
2
11
121
factor }
5}
5 4.84 .
2
25
7
10
}5}
7
10
DE
8
}5}
BC 5 2.8 CD 5 4.2 DE 5 5.6
27. No; Because the triangles are similar, the angle measures
EA
3
7
10
}5}
EA 5 2.1
AB
BC
CD
DE
EA
Fig. 1
3.5
2.8
4.2
5.6
2.1
3
4
28. D; Other leg of nUVW: } 5 }
x
4.5
Fig. 2
5.0
4.0
6.0
8.0
3.0
3x 5 18
b.
are congruent. So, the extended ratio of the angle
measures in nXYZ is x : x 1 30 : 3x.
y
So, the legs of nUVW are 4.5 feet and 6 feet. The
hypotenuse is the longest side, so it must be greater
than 6 feet. The correct answer is D.
29.
1
21
x
Yes, the relationship is linear because the points lie in
a line.
30. Similarity is reflexive, symmetric, and transitive.
Sample answer:
Given: nRAN , nTAG
A
nTAG , nCAB
C
R
7
x
c. Because } 5 }, you know that 10x 5 7y. So, an
4
10
10
10
equation is y 5 }
x. The slope is }
. The slope and
7
7
the scale factor are the same.
34. a.
B
B
A
Sun
N
93,000,000 mi
C
432,500 mi
D
240,000 mi
E
Moon
Earth
b. Sample answer: Because nBDA , nCDE,
T
}
ŽBDA ù ŽCDA, so CE shows that any line of sight
}
from Earth to AB is blocked by the moon. All direct
light from the sun is blocked in this way during a
total eclipse.
G
Reflexive: nRAN , nRAN
Symmetric: nRAN , nTAG, so nTAG , nRAN.
Transitive: nRAN , nTAG and nTAG , nCAB,
so nRAN , nCAB.
c.
CE
ED
DA
AB
240,000
r
}5}
93,000,000
432,500
}5}
103,800,000,000 5 93,000,000r
1116.13 ø r
The radius of the moon is about 1116 miles.
164
Geometry
Worked-Out Solution Key
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
x56
Chapter 6,
continued
OA
3
1
d. } 5 } 5 }
6
2
OC
35. Yes; the images are similar if the original image is a
square. The result will be a square, and all squares are
similar.
21153
1
2
5
10
AB
CD
1
2
}5}5}
38. Let ABCD and FGHJ be similar rectangles.
21153
2
4
8
The ratios are equal, so the corresponding side lengths
are proportional. Because corresponding angles
are congruent and corresponding side lengths are
proportional, n AOB , n COD.
Sample answer:
2
OB
OD
}5}5}
2
The figures are similar squares with a scale factor of }3.
L
36. The ratio of the areas of similar rectangles is the square
K
b
M
a
of the scale factor.
c
d
Q
N
bx
R
Sample answer:
6
4
2
A523458
cx
ax
3
Scale factor: }2
P
18
8
9
4
S
3 2
kFG 1 kGH 1 kHJ 1 kJF
AB 1 BC 1 CD 1 DA
5 }}
So, }}
FG 1 GH 1 HJ 1 JF
FG 1 GH 1 HJ 1 JF
k(FG 1 GH 1 HJ 1 JF)
37. a. The two lines are parallel because they have the same
5 }}
FG 1 GH 1 HJ 1 JF
slope.
5k
b. Ž BOA > Ž DOC by the Vertical Angles Theorem.
Ž OBA > Ž ODC by the Alternate Interior Angles
Theorem.
Ž BAO > Ž DCO by the Alternate Interior Angles
Theorem.
c. Coordinates of A: (x, 0)
y 5 }3 x 2 8
4
0 5 }3 x 2 8
24 5 }3 x
4
8 5 }3 x
23 5 x
65x
Coordinates of B: (0, y)
Coordinates of D: (0, y)
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
0 5 }3 x 1 4
4
y 5 }3 x 1 4
4
AB
5}
.
FG
Because ABCD , FGHJ, you know that
AB
FG
CD
HJ
DA
JF
AB 1 BC 1 CD 1 DA
AB
BC
CD
DA
5}
5}
5}
5}
.
So, }}
HJ
JF
FG 1 GH 1 HJ 1 JF
FG
GH
4
4
BC
GH
} 5 } 5 } 5 }.
Coordinates of C: (x, 0)
4
y 5 }3 x 1 4
AB
The scale factor of ABCD to FGHJ is }
. Let k 5 }
.
FG
FG
5 1 }2 2
3
A 5 3 3 6 5 18
dx
AB
Ratio of Areas: } 5 }
39.
MS
RQ
LM
MR
}5}
x
1
1
x21
}5}
4
x(x 2 1) 5 1
4
y 5 }3 x 2 8
x2 2 x 2 1 5 0
Quadratic Formula:
}
2a
2
4
y 5 }3 (0) 1 4
y 5 }3 (0) 2 8
y54
y 5 28
}
}
26 6 Ïb 2 2 4ac
1 6 Ï5
1 6 Ï1 1 4
}} 5 } 5 }
2
}
1 2 Ï5
because it is a negative number.
You can disregard }
2
}
1 1 Ï5
The coordinates of A, B, C, and D are (23, 0), (0, 4),
(6, 0), and (0, 28), respectively.
The exact value of x is }
.
2
}
1 1 Ï5
}
}
Lengths of sides of n AOB:
MS
1 1 Ï5
2
}5}5}
LM
1
2
OA 5 {0 2 (23){ 5 3
So, PLMS is a golden rectangle.
OB 5 {0 2 4{ 5 4
}}
}
AB 5 Ï(4 2 0)2 1 [0 2 (23)]2 5 Ï25 5 5
Lengths of sides of n COD:
OC 5 {0 2 6{ 5 6
OD 5 {0 2 (28){ 5 8
}}
}
CD 5 Ï(28 2 0)2 1 (0 2 6)2 5 Ï100 5 10
Geometry
Worked-Out Solution Key
165
Chapter 6,
1
x21
}5}
2. H;
1 1 Ï5
}21
11,400 5 285x
2
40 5 x
1
5}
}
Ï5 2 1
}
2
x j dimension on model
190 ft jdimension on actual building
60 in.
285 ft
}5}
1
5}
}
The model of the building is 40 inches wide along the
corresponding front.
}
(Ï5 1 1)
2
5}
+}
}
}
Ï 5 2 1 (Ï 5 1 1)
3. D;
MN
}
2
2
3
12
QS
}
}5}
1 1 Ï5
5}
2
36 5 2(QS)
So, LMRQ is a golden rectangle.
18 5 QS
Perimeter of nQRS 5 QR 1 RS 1 QS
Mixed Review for TAKS
5 15 1 15 1 18
40. A;
5 48
The rate of change is the slope. Choose two points on the
graph: (0, 23) and (2, 0).
y2 2 y1
0 2 (23)
3
5}
5 }2
m5}
x2 2 x1
220
The perimeter of nQRS is 48 centimeters.
4. H;
150
x
1 j number of U.S. dollars
10.63 jnumber of Mexican pesos
}5}
3
The rate of change is }2.
150(10.63) 5 x
1594.5 5 x
41. G;
6
2
4
3
3
1
1
} 5 } inch, } inch, } 5 } inch, } 5 } inch,
32
32
32
16
32
8
16
7
32
1
4
8
32
Kelly has 1594.5 2 640 5 954.5 Mexican pesos left.
5. B;
5
32
} inch, and } 5 } inch. So, the } inch bit is missing.
1,504,000,000 lb
290,000,000 people
}} ø 5.2 pounds per person
42. D;
x
The approximate per capita consumption of peaches in
the U.S. in 2002 was 5.2 pounds per person.
0
y 5 6x 2 17 6(0) 2 17 5 0 2 17 5 217
x
4
y 5 6x 2 17 6(4) 2 17 5 24 2 17 5 7
6.
AC
DF
2
3
AC
6
2
3
}5}
}5}
3(AC) 5 12
AC 5 4
x
7
y 5 6x 2 17 6(7) 2 17 5 42 2 17 5 25
The range of the function is {217, 7, 25}.
Mixed Review for TEKS (p. 380)
1. C;
AB
BC
3
8
426
x
3
8
}5}
}5}
3408 5 3x
1136 5 x
So, BC 5 1136 units and AC 5 AB 1 BC 5
426 1 1136 5 1562 units.
166
10 cm
Scale factor: }
5}
5 }3
15 cm
RS
2(Ï5 1 1)
5}
4
Geometry
Worked-Out Solution Key
AC is 4 units.
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
LM
MR
continued